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2 years ago

What are the conditions that are required for electrical energy to be present in an electrical circuit?

Chemistry

1 answer:

Nat2105 [25]2 years ago

6 0

Answer:

a supply of electric charges which are free to flow, some form of push to move the charges through the circuit and a pathway to carry the charges.

Explanation:

There's your answer have a good day

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Alkali metals, alkaline earth metals, and aluminum all form ions with positive charges equal to the

Amiraneli [1.4K]

B. group number is the correct answer

6 0

3 years ago

Read 2 more answers

The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butano

Lubov Fominskaja [6]

Answer:

Ka = 1.52 E-5

Explanation:

CH3-(CH2)2-COOH ↔ CH3(CH2)2COO- + H3O+

⇒ Ka = [H3O+][CH3)CH2)2COO-] / [CH3(CH2)2COOH]

mass balance:

⇒ C CH3(CH2)2COOH = [CH3(CH2)2COO-] + [CH3(CH2)2COOH] = 1.0 M

charge balance:

⇒ [H3O+] = [CH3(CH2)2COO-]

⇒ Ka = [H3O+]²/(1 - [H3O+])

∴ pH = 2.41 = - Log [H3O+]

⇒ [H3O+] = 3.89 E-3 M

⇒ Ka = (3.89 E-3)² / ( 1 - 3.89 E-3 )

⇒ Ka = 1.519 E-5

3 0

3 years ago

A solution of 2-propanol and 1-octanol behaves ideally. Calculate the chemical potential of 2-propanol in solution relative to t

andrew-mc [135]

Answer:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

Explanation:

The chemical potential of 2-propanol in solution relative to that of pure 2-propanol can be calculated using the following equation:

$\mu (l) = \mu ^{\circ} (l) + R*T*ln(x)$

Where:

μ (l): is the chemical potential of 2-propanol in solution

μ° (l): is the chemical potential of pure 2-propanol

R: is the gas constant = 8.314 J K⁻¹ mol⁻¹

T: is the temperature = 82.3 °C = 355.3 K

x: is the mole fraction of 2-propanol = 0.41

$\mu (l) = \mu ^{\circ} (l) + 8.314 \frac{J}{K*mol}*355.3 K*ln(0.41)$

$\mu (l) = \mu ^{\circ} (l) - 2.63 \cdot 10^{3} J*mol^{-1}$

$\mu (l) - \mu ^{\circ} (l) = - 2.63 \cdot 10^{3} J*mol^{-1}$

Therefore, the chemical potential of 2-propanol in solution relative to that of pure 2-propanol is lower by 2.63x10⁻³.

I hope it helps you!

8 0

3 years ago

Nitrogen dioxide (NO2) cannot be obtained in a pure form in the gas phase because it exists as a mixture of NO2 and N2O4. At 16°

Pavel [41]

Answer:

PNO₂ = 0.49 atm

PN₂O₄ = 0.45 atm

Explanation:

Let's begin with the equation of ideal gas, and derivate from it an equation that involves the density (ρ = m/V).

PV = nRT

n = m/M (m is the mass, and M the molar mass)

$PV = \frac{m}{M}RT$

$PxM = \frac{m}{V}RT$

PxM = ρRT

ρ = PxM/RT

With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

$2.7 = \frac{0.94xMavg}{0.082x289}$

0.94Mavg = 63.9846

Mavg = 68.0687 g/mol

The molar mass of N is 14 g/mol and of O is 16 g/mol, than $M_{NO2} = 46$ g/mol and $M_{N2O4} = 96$ g/mol. Calling y the molar fraction:

$Mavg = M_{NO2}y_{NO2} + M_{N2O4}y_{N2O4}$

And,

$y_{NO2} + y_{N2O4} = 1$

$y_{N2O4} = 1 - y_{NO2}$

So,

$68.0687 = 46y_{NO2} + 92x(1 - y_{NO2})$

$68.0687 - 92 = 46y_{NO2} - 92y_{NO2}$

$46y_{NO2} = 23.9313$

$y_{NO2} = 0.52$

$y_{N2O4} = 0.48$

The partial pressure is the molar fraction multiplied by the total pressure so:

PNO₂ = 0.52x0.94 = 0.49 atm

PN₂O₄ = 0.48x0.94 = 0.45 atm

8 0

3 years ago

The laws of nature:

GenaCL600 [577]

Answer:

Explanation:

they actually say the same thing they differ with smaller things

4 0

1 year ago